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Likurg_2 [28]
3 years ago
14

A lightbulb filled with argon gas has a volume of 75 mL at STP. How many moles of argon gas does the lightbulb contain?

Chemistry
1 answer:
Elan Coil [88]3 years ago
4 0

Answer:

0.00335 moles

Explanation:

From the question, Using

PV = nRT................... Equation 1

Where P = pressure, V = Volume, n = number of moles of argon gas, R = Molar  gas constant, T = Temperature.

make n the subject of the equation

n = PV/RT............... Equation 2

Given: P = 1 atm (standard pressure), T = 273 K (standard temperature), V = 75 mL = 0.075 dm³

Constant: R = 0.082 atm·dm³/K·mol

Substitute into equation 2

n = (1×0.075)/(273×0.082)

n = 0.075/22.386

n = 0.00335 moles

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3 years ago
A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur
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Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

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Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

q=m\times c\times \Delta T

q = heat absorbed by water

m = mass of water = 40.0 g

T_{final} = final temperature of water = 20.0°C

T_{initial = initial temperature of water = 17.0°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

q=40.0\times 4.186\times (20.0-17.0)]

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5 0
3 years ago
Dilution question In many of the experiments, you will be asked to prepare a standard solution by diluting a stock solution. You
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Answer:

We could do two 1:50 dilutions and one 1:4 dilutions.

Explanation:

Hi there!

A solution that is 1000 ug/ ml  (or 1000 mg / l) is 1000 ppm.

Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.

Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.

We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):

First step (1:50 dilution):

Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).

Step 2 (1:50 dilution):

Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)

Step 3 (1:4 dilution):

Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.

6 0
3 years ago
Is NaCl CH3OH LiOH or H2SO4 a bronsted acid
vesna_86 [32]

Answer:

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Bronsted acids are those that donate H+ ions. In this question, H₂SO₄ is a Bronsted acid.

Note: H₂SO₄ is one of seven strong acids that you should try to memorize.

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